What is the equation of locus of |2z+✓z+x-iy|=3 if z=7+24i?

To find the equation of the locus of |2z + √z + x – iy| = 3 when z = 7 + 24i, we can substitute this value into the equation and simplify.

Let’s start with the given equation:

|2z + √z + x – iy| = 3

Now, substitute z = 7 + 24i into the equation:

|2(7 + 24i) + √(7 + 24i) + x – iy| = 3

Next, we’ll simplify the expression inside the absolute value:

2(7 + 24i) + √(7 + 24i) + x – iy

Now, calculate the components separately:

2(7 + 24i) = 14 + 48i

√(7 + 24i) = √(7 + 24i)

So, the expression becomes:

14 + 48i + √(7 + 24i) + x – iy

Now, we want the absolute value of this expression to be equal to 3:

|14 + 48i + √(7 + 24i) + x – iy| = 3

Now, we’ll isolate the real and imaginary parts on one side of the equation and simplify:

|14 + √(7 + 24i) + x| + |48i – y| = 3

This is the equation of the locus. It describes all points (x, y) such that the sum of the absolute values of the real and imaginary parts of the complex number 14 + √(7 + 24i) + x and 48i – y is equal to 3.

Leave a Reply

Your email address will not be published. Required fields are marked *